Question of electric potential

Q: Why the inner surface of the outer sphere has a charge of same magnitude Q?

Since the surface of the inner sphere is given a charge of +Q, it sets up electric field lines, which originate from the charge +Q and end at the the inner surface of the outer sphere.

Electric field lines have the property that they originate and end with the same magnitude of charges, but with opposite polarity. Hence, assuming the hole is small enough, all electric field lines originate from the +Q charge will end at the inner surface of the outer shpere. In other words, the charges on the inner surface of the outer sphere are of the same magnitude as the charges on the inner sphere

With inner sphere positively charged and outer sphere earthed, the outer sphere should carry negative charges.

Potential at any point inside the outer sphere due to its charge should be equal to that at the surface of it, i.e. -Q/(4πε0a)

Potential at the surface inner sphere due to its charge = +Q/(4πε0b)

Hence resultant potential at surface of inner sphere = Q/(4πε0) (1/b - 1-a)

Thus W = qV = Qq/(4πε0) (1/b - 1-a)

2011-02-02 15:48:07 補充：
This is the problem in Gauss's law:

The E-field through a closed surface is equal to the amount of charge enclosed divided by the total surface area.

Since there's no E-field outside the whole sphere, net charge enclosed should be zero and hence the outer sphere's induced charge should be -Q